Construction Is A Piece Of Cake! Right? (Part-3)

Geometry Level 5

Mohit, influenced by Akshat and Anshuman , took a right A B C \bigtriangleup ABC (right angled at B B ), where A B = 5.5 c m AB=5.5 \mathrm{ cm} and B C = 8 c m BC=8 \mathrm{ cm} , and started doing aimless constructions, the steps of which are given below:

( 1 1 ) He drew perpendicular bisector X 1 Y 1 X_1Y_1 of A C AC which intersects A C AC at a point O , O, and another perpendicular bisector X 2 Y 2 X_2Y_2 of A B AB which intersects A B AB at a point G G .

( 2 2 ) Then he constructed a circle taking center O O and radius of the circle as O A OA . Then he constructed an A C E \angle ACE which is equal to A C B \angle ACB such that E E lies on the circle.

( 3 3 ) He then joined B E BE which meets A C AC at F F .

( 4 4 ) Then he joined C G CG which intersects E B EB at H H and X 1 Y 1 X_1Y_1 at J J .

Mohit then wondered what B H O J \frac{BH}{OJ} could possibly be equal to.


The answer is 2.

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2 solutions

Akshay Yadav
Mar 31, 2016

Join B O BO which intersects C G CG at I I

Notice that in A B C \bigtriangleup ABC , B B is the orthocenter, I I is the centroid and O O is the circumcenter.

Hence, B O BO is Euler line .

So B I I O = 2 \frac{BI}{IO}=2 . Also, B I H J I O \bigtriangleup BIH \sim \bigtriangleup JIO , so B H O J = B I I O = 2 \frac{BH}{OJ}=\frac{BI}{IO}=\boxed{2} .

Great solution. Thanks

Ahmad Saad - 5 years, 2 months ago

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Your solution is also awesome, its definitely a new way to solve the problem.

Akshay Yadav - 5 years, 2 months ago
Ahmad Saad
Apr 1, 2016

Another solution

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